How to find the inverse of a function

  • #1
moved into h/w help, so template is missing
T(t) = Ts+(98.6 – Ts)e-kt

rewrite in the form t=g-1(T)

In trying to understand how to find the inverse of this but am having a hard time, please advise.

Thanks,
Kupkake303
 
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Answers and Replies

  • #2
stevendaryl
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This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].
 
  • #3
This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].

So I would put it as 0=Ts+(98.6-Ts)U ??

Doing that would give me -Ts/(98.6-Ts)=U

Then putting the e-kt back in you get -Ts/(98.6-Ts)=e-kt

So then I would then need to get that by it's self so the t is by its self but I can't remember how to do this. Please advise. Thank you
 
  • #4
stevendaryl
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So I would put it as 0=Ts+(98.6-Ts)U ??
No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]
 
  • #5
No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]

Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?
 
  • #6
stevendaryl
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Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?
You want to take the natural log of both sides of the equation.
 
  • #7
You want to take the natural log of both sides of the equation.
okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

so this would be my answer for finding the inverse ..?
 
  • #8
stevendaryl
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okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

so this would be my answer for finding the inverse ..?
Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]
 
  • #9
Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]
Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.
 
  • #10
epenguin
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Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.
 

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